Explanation:
The buoyant force must be greater to float, otherwise it would sink, its like a barrel in water, the more water weight in it the more it sinks, the more air weight the more it rises.
When the spring is extended by 44.5 cm - 34.0 cm = 10.5 cm = 0.105 m, it exerts a restoring force with magnitude R such that the net force on the mass is
∑ F = R - mg = 0
where mg = weight of the mass = (7.00 kg) g = 68.6 N.
It follows that R = 68.6 N, and by Hooke's law, the spring constant is k such that
k (0.105 m) = 68.6 N ⇒ k = (68.6 N) / (0.105 m) ≈ 653 N/m
Answer:
-0.55m/s
Explanation:
Given that: For the boy
Weight = 745N
Velocity = +0.35 m/s
Mass of the boy = ?
g = 9.81m/s^2
W = mg
745 = m×9.81
m = 75.94kg
For the girl
Given that:
Weight = 477 N
g = 9.81m/s^2
m = ?
W = mg
477 = m×9.81m/s^2
m = 48.62kg
To solve for the v of the girl, the two has to add up
48.62kg×v + 75.94kg×+0.35 m/s = 0
48.62v + 26.579 = 0
48.62v = - 26.579
v = -26.579/48.62
v = -0.5466
v = -0.55m/s
Hence, the velocity of the girl is -0.55m/s.
The negative sign is as a result of the two of them moving is opposite direction.
Acceleration of the both masses tied together= 6m/s²
Explanation:
The force is given by F= ma
so 5= m1 (8)
m1=0.625 Kg
for m2
5=m2 (24)
m2=0.208 kg
now total mass= m1+m2=0.625+0.208
Total mass=M=0.833 Kg
now F= ma
5= 0.833 (a)
a= 5/0.833
a=6m/s²
<h3>
Answer:</h3>
1.5 m/s²
<h3>
Explanation:</h3>
We are given;
Force as 60 N
Mass of the Cart as 40 kg
We are required to calculate the acceleration of the cart.
- From the newton's second law of motion, the rate of change in momentum is directly proportional to the resultant force.
- That is, F = ma , where m is the mass and a is the acceleration
Rearranging the formula we can calculate acceleration, a
a = F ÷ m
= 60 N ÷ 40 kg
= 1.5 m/s²
Therefore, the acceleration of the cart is 1.5 m/s²
